Roulette Bernoulli
In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2., 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, you win $35; otherwise you lose $1. Complete parts (a) through (g) below. A)Construct a probability distribution for the. Let us first consider the very simple situation where the fluid is static—that is, v 1 = v 2 = 0. Bernoulli’s equation in that case is. P 1 + ρgh 1 = P 2 + ρgh 2. We can further simplify the equation by taking h 2 = 0 (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative.
For our next problem, let us calculate the probabilty of getting all 38 numbers after spinning a roulette wheel 152 times. (An American roulette wheel has the numbers 1 through 36 plus zero and double zero.) STEP 1 which equals 1.34 × 10240 Then in steps 2, 3, 4 and 5, we will determine how many of those 38152 spins, will contain all 38 numbers. STEP 2 37152 = 2.33 × 10238 36152 = 3.61 × 10236 35152 = 4.99 × 10234 34152 = 6.09 × 10232 33152 = 6.52 × 10230 32152 = 6.06 × 10228 31152 = 4.86 × 10226 30152 = 3.33 × 10224 29152 = 1.93 × 10222 28152 = 9.29 × 10219 27152 = 3.69 × 10217 26152 = 1.19 × 10215 25152 = 3.07 × 10212 24152 = 6.20 × 10209 23152 = 9.61 × 10206 22152 = 1.12 × 10204 21152 = 9.49 × 10200 20152 = 5.71 × 10197 19152 = 2.35 × 10194 18152 = 6.33 × 10190 17152 = 1.07 × 10187 16152 = 1.06 × 10183 15152 = 5.83 × 10178 14152 = 1.63 × 10174 13152 = 2.09 × 10169 12152 = 1.09 × 10164 11152 = 1.96 × 10158 10152 = 1.00 × 10152 9152 = 1.11 × 10145 8152 = 1.86 × 10137 7152 = 2.85 × 10128 6152 = 1.90 × 10118 5152 = 1.75 × 10106 4152 = 3.26 × 1091 3152 = 3.33 × 1072 2152 = 5.71 × 1045 1152 = 1 STEP 3 37 C 38 = 37 36 C 38 = 703 35 C 38 = 8,436 34 C 38 = 73,815 33 C 38 = 501,942 32 C 38 = 2,760,681 31 C 38 = 12,620,256 30 C 38 = 48,903,492 29 C 38 = 163,011,640 28 C 38 = 472,733,756 27 C 38 = 1,203,322,288 26 C 38 = 2,707,475,148 25 C 38 = 5,414,950,296 24 C 38 = 9,669,554,100 23 C 38 = 15,471,286,560 22 C 38 = 22,239,974,430 21 C 38 = 28,781,143,380 20 C 38 = 33,578,000,610 19 C 38 = 35,345,263,800 18 C 38 = 33,578,000,610 17 C 38 = 28,781,143,380 16 C 38 = 22,239,974,430 15 C 38 = 15,471,286,560 14 C 38 = 9,669,554,100 13 C 38 = 5,414,950,296 12 C 38 = 2,707,475,148 11 C 38 = 1,203,322,288 10 C 38 = 472,733,756 9 C 38 = 16,3011,640 8 C 38 = 48,903,492 7 C 38 = 12,620,256 6 C 38 = 2,760,681 5 C 38 = 501,942 4 C 38 = 73,815 3 C 38 = 8,436 2 C 38 = 703 1 C 38 = 38 Basically, this is saying that 37 objects can be chosen from a set of 38 in 37 ways 36 objects can be chosen from a set of 38 in 703 ways ....................................................................................... 2 objects can be chosen from a set of 38 in 703 ways STEP 4 1.34 × 10240 × 1 = 1.34 × 10240 2.33 × 10238 × 37 = 8.84 × 10239 and so on
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Bernoulli distribution Random number distribution that produces bool values according to a Bernoulli distribution, which is described by the following probability mass function: Where the probability of true is p and the probability of false is (1-p). Palace Casino is betting at a Roulette table. He is following a gambling strategy that is often used by prudent gamblers. He has dedicated a capital of $200 to this session with the plan of not winning more than $20. He invariably bets $1 on RED at each spin and plans to do continue playing. Bernoulli trials form one of the simplest yet still most important of all random processes. Ifp= 0 then the gambler always loses and ifp= 1 then the gambler always wins. These trivial cases are not interesting, so we will usually assume that 0.
A Bernoulli Distribution is the probability distribution of a random variable which takes the value 1 with probability p and value 0 with probability 1 – p, i.e.
$$
begin{cases}
1-p & text{for} k=0
p & text{for} k=1
end{cases}$$
We will use the example of left-handedness. Approximately 10% of the population are left-handed (p=0.1).
We want to know, out of a random sample of 10 people, what is the probability of 3 of these 10 people being left handed?
We assign a 1 to each person if they are left handed and 0 otherwise:
- $P(X=1) = 0.1$
- $P(X=0) = 0.9$
A Binomial distribution is derived from the Bernoulli distribution.
We’ll start with the simpler problem:
What is the probability of the first 3 people we pick being left-handed, followed by 7 people being right-handed?
This is just $ 0.1 ^3 times 0.9 ^7$
What if we wanted the last 3 people to be left-handed?
This is just $0.9^7 times 0.1^3$, the same answer.
In fact, no matter how we arrange the 3 people, we will always end up with the same probability ($ 4.7 times 10^{-4} $).
So we have to add up all the ways we can arrange the 3 people being picked.
There are $10!$ ways to arrange 10 people and there are $3!$ ways to arrange the 3 people that are picked and $7!$ ways to arrange the 7 people that aren’t picked.
This is given as:
$$dfrac{10!}{3! 7!}$$
Or more commonly, “10 choose 3”. The “n choose k” notation is written as:
$$
begin{equation*}
binom{n}{k}
end{equation*} = dfrac{n!}{k! (n-k)!}
$$
We can now caclulate the probability that there are 3 left-handed people in a random selection of 10 people as:
$$
P(X=3) = begin{equation*}
binom{10}{3}
end{equation*} (0.1)^3 (0.9)^7
$$
This will generalise such that:
$$
P(X=k) = begin{equation*}
binom{n}{k}
end{equation*} (p)^k (1-p)^{n-k}
$$
Roulette Bernoulli Equation
Scipy’s stats package has a binomial package that can be used to calculate these probabilities:
We can use this function to calculate what the probability of 3 or fewer people being left-handed from a selection of 10 people.
$$
P(X leq 3) = sum_{i=0}^{3} begin{equation*}
binom{10}{i}
end{equation*} (0.1)^i (0.9)^{n-i}
$$
Or we could plot our probability results for each value up to all 10 people being left-handed:
We can see there is almost negligible chance of getting more than 6 left-handed people in a random group of 10 people.
Roulette
On an American roulette wheel there are 38 squares:
- 18 black
- 18 red
- 2 green
We bet on black 10 times in a row, what are the chances of winning more than half of these?
$$
P(X gt 5) = sum_{i=6}^{10} begin{equation*}
binom{10}{i}
end{equation*} bigg(dfrac{18}{38}bigg)^i bigg(1-dfrac{18}{38}bigg)^{n-i}
$$
A Poisson distribution is a limiting version of the binomial distribution, where $n$ becomes large and $np$ approaches some value $lambda$, which is the mean value.
Roulette Bernoulli Game
The Poisson distribution can be used for the number of events in other specified intervals such as distance, area or volume. Examples that may follow a Poisson include the number of phone calls received by a call center per hour and the number of decay events per second from a radioactive source.
It is calculated as:
$ P(k) = e^{-lambda} dfrac{lambda^k}{k!} $
The average number of goals in a World Cup football match is 2.5.
Roulette Bernoulli Games
We would like to know the probability of 4 goals in a match.
Bernoulli Roulette Watch
Again, scipy has in-built functions for calculating this and we can use this to calculate the probability of any number of goals in a World Cup match.